Both pairs of opposite sides parallel — and everything else comes in matched pairs.
Push a rectangular sliding door out of square and the shape that glides open is a parallelogram — both pairs of opposite sides still perfectly parallel, even as the corners lean. That one defining property is generous: it forces a whole cascade of others. The opposite sides turn out to be exactly equal, the opposite angles exactly equal, and the two diagonals cut each other precisely in half. Nothing here is an accident, and nothing needs measuring — we'll prove each fact with a single well-chosen diagonal and the parallel-line angle pairs you already met in transversals and angle pairs. By the end you'll read a parallelogram the way a carpenter reads a frame: name one part and you already know its partner.
A parallelogram is a quadrilateral with both pairs of opposite sides parallel. We write it with the little slanted box, ▱ABCD, naming the vertices in order around the figure, so that
AB ∥ DC and AD ∥ BC.
That is the whole definition — two pairs of parallel sides, nothing more. But watch what the parallels immediately give us. Side AB crosses the two parallel lines AD and BC like a transversal. The angles ∠A and ∠B sit on the same side of that transversal, between the rails — they are co-interior angles, so they are supplementary:
∠A + ∠B = 180°.
Each side of a parallelogram is a transversal of the other pair. So any two consecutive angles are co-interior, hence supplementary. Four corners, two supplementary pairs — already the angles are far from free.
This single observation is the seed of everything below. Before we prove the famous matched-pair properties, let's pin down the supplementary fact with a slider.
Here is the first surprise that the definition hands us for free. Draw one diagonal — say AC — splitting the parallelogram into two triangles, △ABC and △CDA. The diagonal is itself a transversal of each pair of parallel sides, so two pairs of alternate interior angles appear:
∠BAC = ∠DCA (since AB ∥ DC) and ∠BCA = ∠DAC (since AD ∥ BC).
The two triangles also share the side AC. That is one side wedged between two pairs of equal angles — the ASA congruence we met with triangles:
△ABC ≅ △CDA (ASA).
Once the triangles are congruent, their matching parts must be equal. Reading the correspondence A→C, B→D, C→A:
| From the congruence | We read off |
|---|---|
| matching sides | AB = DC and BC = DA |
| matching angle at B and D | ∠B = ∠D |
| add the two half-angles at A and at C | ∠A = ∠C |
So a parallelogram's opposite sides are equal and its opposite angles are equal. The reasoning, in one breath: AB ∥ DC and AD ∥ BC give two pairs of alternate angles; with the shared diagonal AC, △ABC ≅ △CDA by ASA; therefore the opposite sides and opposite angles are equal.
In ▱PQRS, ∠P = 64°. Find the other three angles. Opposite angles are equal, so ∠R = 64°. Consecutive angles are supplementary, so ∠Q = 180° − 64° = 116°, and ∠S = ∠Q = 116°. The four angles 64°, 116°, 64°, 116° add to 360°, just as a quadrilateral must.
Slide to lean the parallelogram from steep to shallow. The side lengths never change, and opposite sides stay equal, opposite angles stay equal — only the lean changes.
Now draw both diagonals, AC and BD, and call their crossing point O. The claim is that O is the exact middle of each diagonal — the diagonals bisect each other.
Look at the two small triangles △AOB and △COD that sit across the crossing from one another. We already know AB = CD (opposite sides, from 16.2.2). Because AB ∥ CD, the diagonal AC makes equal alternate angles ∠OAB = ∠OCD, and the diagonal BD makes equal alternate angles ∠OBA = ∠ODC. That's a side flanked by two equal angles again:
△AOB ≅ △COD (ASA) ⟹ AO = OC and BO = OD.
Each diagonal is cut into two equal halves at O. One careful warning, though: the diagonals are bisected, but they are not equal to each other in a general parallelogram — AC and BD usually have different lengths. (Equal diagonals are the special mark of a rectangle, coming in 16.5.)
Lean the same parallelogram and watch the crossing point O. Each diagonal is always split in half there — even though the two diagonals are different lengths. Flip to "proof" to lift out the congruent triangles.
Two traps to dodge. First, the diagonals bisect each other but they are not equal in general. Second, they do not bisect the angles of the parallelogram — that is the special behavior of a rhombus, not of every parallelogram.
Opposite sides of a parallelogram run like two perfect train rails. Because the rails never tilt toward or away from each other, the perpendicular gap between them is the same everywhere — stand anywhere on the bottom rail, drop a perpendicular to the top rail, and you always measure the same width.
That constant width is the distance between the two parallel lines, and it is the height of the parallelogram that you'll use to compute area later. Notice a parallelogram has two such heights — one between each pair of parallel sides — and they are usually different.
The distance from a point to a line is the length of the perpendicular segment to that line — not a slanted one (recall perpendicular lines). Since opposite sides are parallel, every such perpendicular between them has the same length, so "the distance between two parallel lines" is a single well-defined number.
Start from a single sentence — both pairs of opposite sides parallel — and one well-placed diagonal unlocks the whole figure:
| Property of ▱ABCD | Why |
|---|---|
| Opposite sides parallel: AB ∥ DC, AD ∥ BC | definition |
| Consecutive angles supplementary: ∠A + ∠B = 180° | co-interior angles |
| Opposite sides equal: AB = DC, AD = BC | △ABC ≅ △CDA (ASA) |
| Opposite angles equal: ∠A = ∠C, ∠B = ∠D | △ABC ≅ △CDA (ASA) |
| Diagonals bisect each other: AO = OC, BO = OD | △AOB ≅ △COD (ASA) |
| Diagonals are NOT equal, do NOT bisect angles | those are rectangle / rhombus marks |
Everything comes in matched pairs because the figure itself is built from two matched pairs of sides. In 16.3 we'll read these very properties backward to get tests for proving a quadrilateral is a parallelogram.
In ▱ABCD, ∠A = 110°. Find ∠B, ∠C, and ∠D.
Opposite angles are equal, so ∠C = 110°. Consecutive angles are supplementary, so ∠B = 180° − 110° = 70° and ∠D = ∠B = 70°. Check: 110 + 70 + 110 + 70 = 360°. ✓
A parallelogram has sides AB = 8 and BC = 5. Find its perimeter.
Opposite sides are equal, so DC = AB = 8 and AD = BC = 5. Perimeter = 2(8 + 5) = 26.
The diagonals of ▱ABCD meet at O, and AO = 6. Find the length of diagonal AC.
The diagonals bisect each other, so O is the midpoint of AC and OC = AO = 6. Thus AC = AO + OC = 12. (We cannot find BD from this — the diagonals are not equal.)
In a parallelogram the consecutive angles ∠ABC = 3x and ∠BCD = x. Solve for x and state the four angles.
Consecutive angles are supplementary: 3x + x = 180°, so 4x = 180° and x = 45°. Then ∠BCD = 45°, ∠ABC = 135°, and their opposites repeat: the angles are 135°, 45°, 135°, 45°.
True or false: the diagonals of every parallelogram are equal in length.
False. The diagonals always bisect each other, but they are equal only in the special case of a rectangle. In a leaning parallelogram one diagonal is clearly longer than the other.
Explain why AB = DC in ▱ABCD, in one reasoned line.
Draw diagonal AC. Since AB ∥ DC and AD ∥ BC, alternate angles give ∠BAC = ∠DCA and ∠BCA = ∠DAC; with shared side AC, △ABC ≅ △CDA (ASA), so the matching sides give AB = DC.
Six questions to lock it in. Tap the answer you think is right.
This lesson proves the three classical properties of a parallelogram — opposite sides equal, opposite angles equal, and diagonals bisecting each other — each from the single definition using one diagonal and the parallel-line angle pairs from Stage 14. The recurring move is worth naming aloud with students: a diagonal turns "two parallel lines + a transversal" into a triangle congruence (ASA), and congruent triangles hand back equal parts. Encourage students to state the reason at every step, not just the result; the chain "alternate angles ⇒ ASA ⇒ equal parts" is the template they'll reuse for the rest of plane geometry.
The misconception to watch for is about the diagonals. Many students over-remember "the diagonals do something nice" and wrongly conclude the diagonals are equal (true only for a rectangle) or that they bisect the angles (true only for a rhombus). For a general parallelogram the diagonals only bisect each other. The shear sliders are built precisely to make this visible: as the figure leans, the diagonals are obviously unequal yet are still cut exactly in half at O. A good prompt: "Drag it until the diagonals look equal — what shape did you just make?"
Common Core alignment. This is core HS G-CO.C.11 — prove theorems about parallelograms (opposite sides and angles are congruent; the diagonals bisect each other) — building on the angle and congruence work of 8.G. The perpendicular-distance idea in 16.2.4 prepares the area formula (base × height) and previews the height used throughout mensuration.